Kinematics in One Dimension
Chapter 2
Chapter 2: Kinematics in One Dimension
• Objectives
• Displacement, speed, velocity and
acceleration
• Equations of kinematics and
applications
• Graphical analysis of velocity and
acceleration
• Examples
Objectives
¾ Introduce displacement and discuss the difference
between displacement and distance.
¾ Introduce speed, velocity (average, instantaneous,
difference between speed and velocity).
¾ Introduce acceleration (average, instantaneous)
and illustrate the difference between velocity and
acceleration.
¾ Introduce and discuss the correct usage of the
equations of kinematics in one
dimension(examples, problem solving strategies).
¾ Describe free fall and introduce the acceleration
due to gravity.
Displacement, Speed, Velocity and
Acceleration
Motion in one dimension
Displacement, Speed, Velocity and
Acceleration
Displacement
Origin
t0
x0
Elapsed time: ∆t = t –t0
Displacement: ∆x = x - x0
x
Note: Displacement is a vector
t
Displacement, Speed, Velocity and
Acceleration
Speed and velocity
m
m/s
Average Speed =
Distance
Elapsed time
s
Displacement
Average velocity =
Elapsed time
x – x0
v=
= ∆x
∆t
t – t0
∆x = dx
Instantaneous velocity v =∆lim
t 0 ∆t
dt
Displacement, Speed, Velocity and
Acceleration
Acceleration
m/s
m/s2
Average acceleration =
Change in velocity
Elapsed time
v – v0
a=
= ∆v
∆t
t – t0
Instantaneous acceleration
∆v = dv
a =∆lim
t 0 ∆t
dt
s
Equations of Kinematics and
Applications
Concept
Equations of Kinematics and
Applications
Motion with constant acceleration
v = v0 + at
x = ½ (v0 +v)t
2
x = x0+v0t + ½ at
v2 = v02 + 2a(x-x0)
Example 1
Two soccer players start from rest, 48 m apart.
They run directly toward each other. The
magnitude of their accelerations are
a1=0.50m/s2 and a2=0.30m/s2 respectively.
(a) How much time passes before they
collide?
(b) At the instant they collide, how far has the
first player run?
Solution:
(a) t = 11s; (b) x1 = 3.0 101 m
Equations of Kinematics and
Applications
Freely Falling Bodies
g = 9.80 m/s2
g = 32.2 ft/s2
Example 2
A ball is thrown upward from the top of a
25.0 m tall building. The ball’s initial speed
is 12.0 m/s. At the same instant, a person is
running on the ground at a distance of 31.0
m from the building. What must be the
average speed of the person if he is to catch
the ball at the bottom of the building?
Solution:
t = 3.79 s; v=x/t = 8.18 m/s
Graphical Analysis of Velocity
and Acceleration
Graphical Analysis of Velocity
and Acceleration
Example 3
A person who walks for exercise
produces the following positiontime graph. (a) without doing any
calculations,
decide
which
segments of the graph (A, B, C,
or D) indicate positive, negative,
and zero average velocities. (b)
Calculate the average velocity
for each segment.
Solution:
(a) Positive for A and C; negative for B and zero for D
(b) vA = 6.25 km/h, vB= -3.75 km/h, vC = 0.625km/h, vD = 0 km/h